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polished crystal structures a bit

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docs/9_crystal_structure.md
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...@@ -42,24 +42,25 @@ _based on chapter 12 of the book_ ...@@ -42,24 +42,25 @@ _based on chapter 12 of the book_
After this lecture you will be able to: After this lecture you will be able to:
- Describe any crystal using crystallographic terminology, and interpret this terminology - Describe any crystal using crystallographic terminology, and interpret this terminology
- Compute the volume filling fraction given a crystal structure - Compute the volume filling factor given a crystal structure
- Determine the primitive, conventional, and Wigner-Seitz unit cells of a given lattice - Determine the primitive, conventional, and Wigner-Seitz unit cells of a given lattice
- Determine the Miller planes of a given lattice - Determine the Miller planes of a given lattice
## Crystal classification ## Crystal classification
In the past few lectures, we derived some very important physical quantities for phonons and electrons, such as the effective mass, the dispersion relation.
These systems we considered were mainly 1D. In the past few lectures, we derived some very important physical quantities for phonons and electrons, such as the effective mass and the dispersion relation.
The systems we considered were mainly 1D.
But most solids, such as crystals, are 3D structures. But most solids, such as crystals, are 3D structures.
Describing 3D system is not much harder than describing 1D systems. Describing 3D systems is not much harder than describing 1D systems.
It does however, require a new language and framework in order to fully describe such structures. It does however, require a new language and framework to fully describe such structures.
Therefore the upcoming two lectures will focus on developing, understanding, and applying this language and framework. Therefore the upcoming two lectures will focus on developing, understanding, and applying this language and framework.
### Lattices and unit cells ### Lattices and unit cells
Most of solid state physics deals with crystals, which are periodic multi-atomic structures. Most of solid state physics deals with crystals, which are periodic multi-atomic structures.
To describe such periodic structures, we need a simple framework. To describe such periodic structures, we need a framework.
Such a framework is given by the concept of a **lattice**: Such a framework is given by the concept of a **lattice**:
> A lattice is an infinite set of points defined by an integer sums of a set of linearly independent **primitive lattice vectors**(will be explained later on). > A lattice is an infinite set of points defined by integer sums of a set of linearly independent **primitive lattice vectors**(will be explained later on).
Which for a 3D system translates to: Which for a 3D system translates to:
...@@ -74,11 +75,11 @@ This definition is pretty rigorous. ...@@ -74,11 +75,11 @@ This definition is pretty rigorous.
But there exist multiple equivalent definitions of a lattice. But there exist multiple equivalent definitions of a lattice.
A more informal definition is: A more informal definition is:
> A lattice is a set of points where the environment of each point is the same. > A lattice is a set of points, where the environment of each point is the same.
In the image below we show a 2D *simple square lattice* (panel A). In the image below we show a 2D *simple square lattice* (panel A).
Each of the black dots are called lattice points. The black dots are the lattice points.
Because the environment of each lattice point is the same, this configuration of lattice points forms a proper lattice. Because the environment of each point is the same, this set of points forms a lattice.
```python ```python
# Define the lattice vectors # Define the lattice vectors
...@@ -219,25 +220,24 @@ A vector connecting any two lattice points is called a **lattice vector**. ...@@ -219,25 +220,24 @@ A vector connecting any two lattice points is called a **lattice vector**.
Suppose we choose two linearly independent lattice vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ (panel B). Suppose we choose two linearly independent lattice vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ (panel B).
These two lattice vectors span an area which is called a **unit cell**: These two lattice vectors span an area which is called a **unit cell**:
> A unit cell is a region of space such that when many identical units are stacked together it tiles (completely fills) all of space and reconstructs the full structure. > A unit cell is a spatial region that, when many identical unit cells are stacked together, tiles (completely fills) all of space and reconstructs the full structure.
In the case of a 3D lattice, we need to choose three linearly independent lattice vectors and thus the unit cell will be a volume instead of an area. In the case of a 3D lattice, we need to choose three linearly independent lattice vectors. Therefore, the unit cell will be a volume instead of an area.
If the chosen unit cell only contains a single lattice point, we speak of a **primitive unit cell**. If the chosen unit cell only contains a single lattice point, we speak of a **primitive unit cell**.
The lattice vectors which construct the primitive unit cell are called **primitive lattice vectors**. The lattice vectors which span a primitive unit cell are called **primitive lattice vectors**.
Because the primitive unit cell is constructed out of a set of linearly independent primitive lattice vectors, the primitive unit cell can be repeated infinitely many times to map out the entire lattice! Because the primitive unit cell is constructed from a set of linearly independent, primitive lattice vectors, it can be repeated infinitely many times to map out the entire lattice!
The unit cell, which was conveniently chosen in panel B, is such a primitive unit cell. The unit cell, which was conveniently chosen in panel B of the figure above, is a primitive unit cell because it contains only one lattice point.
At first glance it might seem that there are four lattice point inside the primitive unit cell instead of one. At first glance, it might seem that there are four lattice points inside the cell, but each point lies only for $\frac{1}{4}$'th inside the cell.
However, each point only occupies the lattice by $\frac{1}{4}$'th. Therefore, there is exactly $4 \times \frac{1}{4}=1$ lattice point in the unit cell and thus it is primitive!
Thus there is exactly $4 \times \frac{1}{4}=1$ lattice point in the unit cell and is thus it is primitive!
The choice of the primitive lattice vectors is **not unique**. The choice of primitive lattice vectors is **not unique**.
In panel C, we show the same lattice plotted with two different unit cells. In panel C, we show the same lattice plotted with two different unit cells.
Both choices are primitive unit cells and thus make it possible to map out the entire lattice! Both choices are primitive unit cells and can therefore be used to construct the entire lattice!
??? Question "Examine the primitive unit cell spanned by the lattice vectors $\tilde{\mathbf{a}}_1$ and $\tilde{\mathbf{a}}_2$. How much does each lattice point occupy the unit cell?" ??? Question "Examine the primitive unit cell spanned by the lattice vectors $\tilde{\mathbf{a}}_1$ and $\tilde{\mathbf{a}}_2$. What fraction of each lattice point lies within the unit cell?"
Two points occupy it for 1/8'th and two points for 3/8'th. Two points lie for 1/8'th inside the unit cell, and the other two points for 3/8'th.
### Periodic structures. ### Periodic structures.
Equipped with the basic definitions of a lattice, (primitive) lattice vectors, and (primitive) unit cells, we now apply our knowledge to an actual periodic structure. Equipped with the basic definitions of a lattice, (primitive) lattice vectors, and (primitive) unit cells, we now apply our knowledge to an actual periodic structure.
...@@ -342,9 +342,9 @@ layout = dict( ...@@ -342,9 +342,9 @@ layout = dict(
go.Figure(data=data, layout=layout) go.Figure(data=data, layout=layout)
``` ```
There are several ways to assign lattice points to the periodic structure. There are many reasonable ways to assign lattice points to the periodic structure.
We can assign the lattice points to the stars themselves. We can for instance assign the lattice points to the stars themselves.
This is a valid choice of a lattice for the periodic structure because each lattice point has the same environment. This set of points indeed defines a lattice, because each of the points has the same environment.
Because the lattice points form triangles, this lattice is called a **triangular lattice**. Because the lattice points form triangles, this lattice is called a **triangular lattice**.
The choice of a lattice also defines two linearly independent primitive lattice vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ (panel B): The choice of a lattice also defines two linearly independent primitive lattice vectors $\mathbf{a}_1$ and $\mathbf{a}_2$ (panel B):
...@@ -353,7 +353,7 @@ $$ ...@@ -353,7 +353,7 @@ $$
\mathbf{a}_1 = \hat{\mathbf{x}} = \left[ 1, 0\right], \quad \mathbf{a}_2 = \frac{1}{2}\hat{\mathbf{x}} + \frac{\sqrt{3}}{2} \hat{\mathbf{y}}= \left[1/2, \sqrt{3}/2\right]. \mathbf{a}_1 = \hat{\mathbf{x}} = \left[ 1, 0\right], \quad \mathbf{a}_2 = \frac{1}{2}\hat{\mathbf{x}} + \frac{\sqrt{3}}{2} \hat{\mathbf{y}}= \left[1/2, \sqrt{3}/2\right].
$$ $$
With these primitive lattice vector, the lattice is given by With these primitive lattice vectors, the lattice is given by
$$ $$
\mathbf{R}_{\left[n_{1}, n_{2}\right]}=n_{1} \left[ 1, 0 \right]+n_{2} \mathbf{R}_{\left[n_{1}, n_{2}\right]}=n_{1} \left[ 1, 0 \right]+n_{2}
...@@ -362,12 +362,12 @@ $$ ...@@ -362,12 +362,12 @@ $$
However, our description so far is insufficient to describe the periodic structure. However, our description so far is insufficient to describe the periodic structure.
Although we mapped out the entire lattice, we still do not have any information about the periodic star-like structure itself. Although we mapped out the entire lattice, we still do not have any information about the periodic star-like structure itself.
To include the information of the periodic structure, we need to define a **basis** (do not confuse this with the definition of a basis in linear algebra): To include the information of the periodic structure, we need to define a **basis** (do not confuse this with the definition of a basis in linear algebra):
> The description of objects with respect to the reference lattice point is known as a basis. > The description of objects with respect to the reference lattice point is known as a basis.
The reference lattice point is the chosen lattice point to which we apply the lattice vectors in order to reconstruct the lattice. The reference lattice point is the chosen lattice point to which we apply the lattice vectors in order to reconstruct the lattice.
In our case, we chose the reference point as $[0, 0]$. In our case, we chose the reference point at $[0, 0]$.
With respect to the reference point, the star is located at $[0, 0]$. With respect to the reference point, the star is located at $[0, 0]$.
The location of all the stars in the lattice with respect to the reference lattice point is then given by: The location of all the stars in the lattice with respect to the reference lattice point is then given by:
...@@ -394,16 +394,14 @@ In our case, the basis is $\star(0,0) = 0 \mathbf{a}_1 +0 \mathbf{a}_2$. ...@@ -394,16 +394,14 @@ In our case, the basis is $\star(0,0) = 0 \mathbf{a}_1 +0 \mathbf{a}_2$.
Here $\star$ means the type of object that is specified by the basis. Here $\star$ means the type of object that is specified by the basis.
E.g. if we were to have carbon atoms (C) as well, we would write that down as $\mathrm{C}(f_1, f_2)$. E.g. if we were to have carbon atoms (C) as well, we would write that down as $\mathrm{C}(f_1, f_2)$.
Similar to primitive lattice vectors, the choice of a lattice **not unique**. Similar to the choice of primitive lattice vectors, the choice of a lattice is **not unique**.
In panel C we show the same periodic structure, but now with the lattice translated by $1/2 \mathbf{a}_1$. In panel C we show the same periodic structure, but now with the lattice points translated by $1/2 \mathbf{a}_1$.
This choice still fulfills the definition of a lattice: the environment of each lattice point is the same. This set of points still fulfills the definition of a lattice because the environment of each point is the same.
The lattice is only translated and thus we keep using the same primitive lattice vectors. Since the lattice is only translated, we can use the same primitive lattice vectors.
However, we must change the basis: the stars are not located on the lattice points anymore, but are now located at $[1/2,0]$, which is half of $\mathbf{a}_1$. However, we must change the basis: the stars are now not located on the lattice points anymore. Instead, they are located at $1/2 \mathbf{a}_{1}+0\mathbf{a}_{2}$ with respect to the reference lattice point. Therefore, the basis is $\star(1/2, 0)$.
Therefore, the basis is given by: $\star(1/2, 0)$.
As a reminder, this implies that the position of the star is at $1/2 \mathbf{a}_{1}+0\mathbf{a}_{2}$ with respect to the reference lattice point.
The location of the stars throughout the periodic structure with respect to the reference lattice point is then described by The location of the stars with respect to the reference lattice point is then described by
$$ $$
\mathbf{R}_{\left[n_{1}, n_{2}\right]}^{\mathrm{star}} = \left[1/2, 0 \right] + n_{1} \left[ 1, 0 \right]+n_{2} \mathbf{R}_{\left[n_{1}, n_{2}\right]}^{\mathrm{star}} = \left[1/2, 0 \right] + n_{1} \left[ 1, 0 \right]+n_{2}
...@@ -411,12 +409,12 @@ $$ ...@@ -411,12 +409,12 @@ $$
$$ $$
### Conventional unit cell ### Conventional unit cell
Sometimes the primitive unit cell might not be the most practical choice because it is often non-orthogonal. Sometimes the primitive unit cell is not the most practical choice, because it is often non-orthogonal.
However, we can define a **conventional unit cell** with an orthogonal set of lattice vectors (panel D). However, we can define a **conventional unit cell** with an orthogonal set of lattice vectors (panel D).
Conventional unit cells can contain multiple lattice points, as shown in the plot: since there is an additional lattice point at the center, there are $1 + \frac{1}{4} \times 4 = 2 $ lattice points in the conventional unit cell. Conventional unit cells can contain multiple lattice points, as shown in the plot: since there is an additional lattice point at the center, there are $1 + \frac{1}{4} \times 4 = 2 $ lattice points in the conventional unit cell.
There is a slight problem with this definition of the lattice vectors: no integer linear combination of lattice vectors is able to produce the center lattice points in the unit cell. There is a slight problem with this definition of the lattice vectors: no integer linear combination of lattice vectors is able to produce the center lattice points in the unit cell.
In order to map out the entire lattice, we need to include an extra star/lattice point in the basis. In order to map out the entire lattice, we need to include an extra star in the basis.
Since there is one star at the corner of the unit cell and one in the centre, the basis is: $\star = (0,0),(1/2,1/2)$. Since there is one star at the corner of the unit cell and one in the centre, the basis is: $\star = (0,0),(1/2,1/2)$.
The corresponding locations of the stars with respect to the reference lattice point are then given by: The corresponding locations of the stars with respect to the reference lattice point are then given by:
...@@ -434,21 +432,20 @@ Alternatively, one can say that the complete crystal structure is made up from t ...@@ -434,21 +432,20 @@ Alternatively, one can say that the complete crystal structure is made up from t
### General procedure to analyze a periodic structure ### General procedure to analyze a periodic structure
In order to summarize what we explained above, we give a short procedure on how to analyze periodic structures. To summarize what we explained above, we give a short procedure on how to analyze periodic structures:
The procedure is as follows:
1. Choose origin (can be an atom, not necessary) 1. Choose a point as the origin (this can be an atom, but does not have to be)
2. Find other lattice points that are identical 2. Find other points for which the environment is identical to that of the first point.
3. Choose lattice vectors that translate between these lattice points: either primitive or not primitive (in which case you require an additional basis point) 3. Choose lattice vectors that connect these lattice points: either primitive or not primitive (in the latter case you need an additional basis point)
- lengths of lattice vectors and angle(s) between them fully define the crystal lattice 4. Specify the basis
4. Specify basis
With the definition of lattice and a correct basis, the location of every atom in the periodic structure is known and the crystal can be reconstructed. With the definition of the lattice and the basis, the location of every atom in the periodic structure is known and the crystal can be reconstructed. To recall: a crystal structure = lattice + basis
### Example: analyzing the graphene crystal structure ### Example: analyzing the graphene crystal structure
Let us apply our knowledge to an actual physical periodic structure: graphene.
Graphene is made out of a single layer of carbon atoms arranged in a honeycomb shape. Let us apply our knowledge to an actual crystal structure, namely that of graphene.
The nearest neighbor interatomic distance is $a$. Graphene consists of a single layer of carbon atoms arranged in a honeycomb shape.
The nearest-neighbour interatomic distance is $a$.
Below we show the crystal structure of graphene (panel A). Below we show the crystal structure of graphene (panel A).
```python ```python
...@@ -609,43 +606,36 @@ go.Figure(data=data, layout=layout) ...@@ -609,43 +606,36 @@ go.Figure(data=data, layout=layout)
``` ```
Our first task is to find suitable lattice points. Our first task is to find suitable lattice points.
We start by choosing a lattice point on the $(0,0)$ coordinate. We start by choosing a lattice point at $(x,y)=(0,0)$.
We see that not all carbon atoms have the same environment with respect to our chosen lattice point. Crucially, we see that not all atoms have the same environment.
Hence, not all carbon atoms coincide with lattice points! Hence, the set of all carbon atoms do not form a lattice!
Only those with the same environment are valid lattice points. Only those with the same environment are valid lattice points.
In panel B we show the lattice of graphene. In panel B we show the lattice of graphene.
Because primitive lattice vectors are not unique, we are free to choose them. Primitive lattice vectors are not unique. In panel C, a possible choice of primitive lattice vectors is shown.
In panel C a choice of primitive lattice vectors is shown. Using geometry, we find
With a little help from geometry, we find that $\mathbf{a}_1$ and $\mathbf{a}_2$ are
$$ $$
\mathbf{a}_1 = [\sqrt{3}, 0] a, \quad \mathbf{a}_2 = [\sqrt{3}/2, 3/2] a. \mathbf{a}_1 = [\sqrt{3}, 0] a, \quad \mathbf{a}_2 = [\sqrt{3}/2, 3/2] a.
$$ $$
Where $a$ is the interatomic distance. Having found the lattice vectors, the only thing left is to specify the basis. We observe that each primitive unit cell contains two carbon atoms: one at $\mathbf{r}_0 = (0,0)$ and the other at $\mathbf{r}_1 = (\sqrt{3},1)a$.
With these lattice vectors, we are able to map out the entire lattice. We want to express these coordinates as fractional coordinates of our lattice vectors.
The only thing that is left for us is to specify the basis. For the atom at $\mathbf{r}_0$ this is easy: $\mathrm{C}(0,0)$.
Each primtive unit cell contains two carbon atoms. (here C stand for carbon).
One at the reference point $(0,0)$ and the other at the location $(\sqrt{3},1)a$. For the other atom, we use
We want to express these coordinates as fractional coordinates of the chosen lattice vectors.
For the atom at the reference point this is easy: $\mathrm{C}(0,0)$. $$
Here C stand for carbon. \mathbf{r_1} = f_1\mathbf{a}_1 + f_2\mathbf{a}_2
$$
In order to find the fractional coordinates of the second carbon atom, it is convenient to first look at the $y$ coordinate of the atom, which is $a$.
Because only $\mathbf{a}_2$ has a nonzero $y$ component, we easily find the fractional coordinate of $\mathbf{a}_2$. We first look at the $y$-component of $\mathbf{r}_1$, which is $a$. Because only $\mathbf{a}_2$ has a nonzero $y$ component, we conclude $f_2=2/3$. We then look at the $x$-component of $\mathbf{r}_1$, which is $\sqrt{3}a$. It should satisfy
It simply is $\frac{a}{3a/2} = 2/3$.
To find the fractional coordinate of $\mathbf{a}_1$, we use the fractional coordinate of $\mathbf{a}_2$.
Multiplying the $x$ component of $\mathbf{a}_2$ by $2/3$ yields $1/\sqrt{3}$.
We know that
\begin{align} \begin{align}
\sqrt{3} &= f_1 a_{1,x} + f_2 a_{2,x}\\ \sqrt{3}a &= f_1 a_{1,x} + f_2 a_{2,x}\\
&= f_1 \sqrt{3} + 1/\sqrt{3},
\end{align} \end{align}
Bringing $1/\sqrt{3}$ to the other side and dividing both sides by $\sqrt{3}$ yields $f_1 = 2/3$. which we solve to find $f_1 = 2/3$. Hence, the fractional coordinates of the second atom are $\mathrm{C}(2/3, 2/3)$.
Hence the basis of the second atom is $\mathrm{C}(2/3, 2/3)$.
In panel D we show the conventional unit cell with the lattice vectors In panel D we show the conventional unit cell with the lattice vectors
...@@ -710,7 +700,7 @@ Thus, the conventional unit cell contains $8\times 1/8+1 = 2$ lattice points and ...@@ -710,7 +700,7 @@ Thus, the conventional unit cell contains $8\times 1/8+1 = 2$ lattice points and
In the exercises you will figure out a proper set of primitive lattice vectors for the bcc lattice. In the exercises you will figure out a proper set of primitive lattice vectors for the bcc lattice.
### Face centered cubic lattice ### Face-centered cubic lattice
If we add an additional point to the middle of every face of the simple cubic cell, we obtain the **face-centered cubic** (**fcc**) lattice (panel C). If we add an additional point to the middle of every face of the simple cubic cell, we obtain the **face-centered cubic** (**fcc**) lattice (panel C).
There are is $1/2$ of a lattice point at each face inside the lattice in addition to the corner lattice point. There are is $1/2$ of a lattice point at each face inside the lattice in addition to the corner lattice point.
Hence there are a total of $8 \times 1/8 + 6\times 1/2 = 4$ lattice points inside the unit cell and thus it is not primitive. Hence there are a total of $8 \times 1/8 + 6\times 1/2 = 4$ lattice points inside the unit cell and thus it is not primitive.
...@@ -1067,26 +1057,22 @@ fig.update_layout( ...@@ -1067,26 +1057,22 @@ fig.update_layout(
fig fig
``` ```
We first need to blow up the atoms until they touch each other. We first need to blow up the atoms until they touch each other (panel B). We see that on the diagonal, the atoms touch each other.
Thankfully, the faces of the fcc are rotationally symmetric and we obtain that all four corner atoms touch the centre atom (panel B). Because the atoms have radius $R$, the length of the diagonal is equal to $4R$.
We see that on the diagonal, the atoms touch eachother. But, as previously stated, the sides of the unit cell have length $a$. Therefore, the diagonal of the unit cell is also equal to $\sqrt{2}a$.
Because the atoms have a radius $R$, the length of the diagonal should be equal to $4R$.
But, as previously stated, the sides of the unit cell are of length $a$ and thus the diagonal of the unit cell should also be equal to $\sqrt{2}a$.
This implies that This implies that
$$ $$
R = \frac{a}{2\sqrt{2}}. R = \frac{a}{2\sqrt{2}}.
$$ $$
With the deduced radius of the atom, we easily calculate $V_{\mathrm{atom}}$: Having expressed the radius of the atom in terms of the lattice constant, the volume of the atom $V_{\mathrm{atom}}$ can be expressed as:
$$ $$
V_{\mathrm{atom}} = \frac{4\pi}{3}R^3 = \frac{4\pi}{3} \left( \frac{a}{2\sqrt{2}} \right)^3 = \frac{\pi a^3}{12 \sqrt{2}}. V_{\mathrm{atom}} = \frac{4\pi}{3}R^3 = \frac{4\pi}{3} \left( \frac{a}{2\sqrt{2}} \right)^3 = \frac{\pi a^3}{12 \sqrt{2}}.
$$ $$
The only thing that is left for us to determine is the number of atoms in the unit cell. We showed earlier that the number of atoms in the fcc unit cell is 4. Therefore, we now have all the information to calculate the filling factor:
We showed earlier that the number of atoms in the fcc unit cell is 4.
We now have all the information we need to calculate the filling factor:
\begin{align} \begin{align}
F &= \frac{ N_{\mathrm{atom}} V_{\mathrm{atom}} }{V_{\mathrm{cell}}}\\ F &= \frac{ N_{\mathrm{atom}} V_{\mathrm{atom}} }{V_{\mathrm{cell}}}\\
...@@ -1101,13 +1087,12 @@ The packing limit was theorized by Kepler in 1571–1630 and proven by Hales _et ...@@ -1101,13 +1087,12 @@ The packing limit was theorized by Kepler in 1571–1630 and proven by Hales _et
In the exercises you will also calculate the filling factor of the bcc lattice. In the exercises you will also calculate the filling factor of the bcc lattice.
### Miller planes ### Miller planes
When fabricating crystals it is important to know both the orientation and the surface of the crystal. When fabricating crystals it is important to know both the orientation and the surface of the crystal.
Different cuts of a crystal lead to different surfaces. Different cuts of a crystal lead to different surfaces.
In the chemical industry, this is especially significant because different surfaces lead to different chemical properties and thus is one of the foundations of research in catalysts. In the chemical industry, this is especially significant because different surfaces lead to different chemical properties and thus is one of the foundations of research in catalysts.
Therefore, we seek a way to describe different planes of a crystal within our developed framework. Therefore, we seek a way to describe the different planes of a crystal. This leads us to a very important concept - **Miller planes**.
This leads us to a very important concept - **Miller planes**.
To explain Miller planes, let's start off with a simple cubic lattice: To explain Miller planes, let's start off with a simple cubic lattice:
![](figures/cubic_mod.svg) ![](figures/cubic_mod.svg)
...@@ -1116,38 +1101,38 @@ Where $|{\bf a}_1|=|{\bf a}_2|=|{\bf a}_3|\equiv a$. ...@@ -1116,38 +1101,38 @@ Where $|{\bf a}_1|=|{\bf a}_2|=|{\bf a}_3|\equiv a$.
We can cut multiple planes through the cubic lattice. We can cut multiple planes through the cubic lattice.
Miller planes describe such planes with a set of indices. Miller planes describe such planes with a set of indices.
The plane designated by Miller indices $(u,v,w)$ intersects lattice vector ${\bf a}_1$ at $\frac{|{\bf a}_1|}{u}$, ${\bf a}_2$ at $\frac{|{\bf a}_2|}{v}$ and ${\bf a}_3$ at $\frac{|{\bf a}_3|}{w}$. The plane defined by Miller indices $(u,v,w)$ intersects lattice vector ${\bf a}_1$ at $\frac{|{\bf a}_1|}{u}$, ${\bf a}_2$ at $\frac{|{\bf a}_2|}{v}$ and ${\bf a}_3$ at $\frac{|{\bf a}_3|}{w}$.
![](figures/miller.svg) ![](figures/miller.svg)
Miller index 0 means that the plane is parallel to that axis (intersection at "$\frac{|{\bf a}_3|}{0}\rightarrow\infty$"). A bar above a Miller index means intersection at a negative coordinate. A Miller index equal to 0 means that the plane is parallel to that axis (intersection at "$\frac{|{\bf a}_3|}{0}\rightarrow\infty$"). A bar above a Miller index means intersection at a negative coordinate.
If a crystal is symmetric under $90^\circ$ rotations, then $(100)$, $(010)$ and $(001)$ are physically indistinguishable. If a crystal is symmetric under $90^\circ$ rotations, then $(100)$, $(010)$ and $(001)$ are physically indistinguishable.
Therefore, we use the notation $\{100\}$ to indicate a whole family of these symmetry-related planes. Therefore, we use the notation $\{100\}$ to indicate a whole family of these symmetry-related planes.
In a cubic crystal, $[100]$ (this is a vector) is perpendicular to $(100)$ $\rightarrow$ proof in problem set. In a cubic crystal, $[100]$ (this is a vector) is perpendicular to $(100)$ $\rightarrow$ proof in problem set.
Why are these Miller planes usefull? Why are these Miller planes useful?
It allows us to know the exact orientation of a crystal structure if the crystal structure is known. It enables us to know the exact orientation of a crystal structure if the crystal structure is known.
## Summary ## Summary
* We discussed how a crystal is constructed through the definition of a lattice and a proper basis.
* A crystal is constructed through the definition of a lattice and a basis.
* We introduced several important concepts that allow us to describe crystal structure: lattice, lattice vectors, basis, primitive & conventional unit cells, and Miller planes. * We introduced several important concepts that allow us to describe crystal structure: lattice, lattice vectors, basis, primitive & conventional unit cells, and Miller planes.
* We introduced several common 3D lattices: simple-cubic, FCC, and BCC. * We introduced several common 3D lattices: simple-cubic, FCC, and BCC.
* We discussed how to compute the filling factor. * We discussed how to compute the filling factor.
## Exercises ## Exercises
### Quick warm-up exercises ### Warm-up exercises
1. State the definition of a primitive unit cell.
What can be said about its volume? 1. State the definition of a primitive unit cell.
2. Draw the conventional unit cell of a FCC and the BCC. 2. Suppose a conventional unit cell contains 4 lattice points and has volume $a^3$, what would be the volume of the primitive unit cell?
Write down the primitive lattice vectors and the basis of each lattice. 3. Draw the conventional unit cell of the FCC and the BCC lattice. Write down the associated basis vectors. In addition, find a set of primitive lattice vectors.
3. Suppose you find the primitive unit cell of a diatomic crystal. 4. Suppose you find the primitive unit cell of a diatomic crystal.
How many basis vectors do you minimally need to describe the crystal? How many basis vectors do you minimally need to describe the crystal?
Can a diatomic crystal require more basis vectors? Can a diatomic crystal require more basis vectors?
4. Calculate the filling factor of a simple cubic lattice. 5. Calculate the filling factor of a simple cubic lattice.
5. Sketch the $(110),(1\bar{1}0),(111)$ miller planes of a simple cubic lattice. 6. Sketch the $(110),(1\bar{1}0),(111)$ Miller planes of a simple cubic lattice.
### Exercise 1: Diatomic crystal ### Exercise 1: Diatomic crystal
...@@ -1163,21 +1148,16 @@ plt.plot(x,y,'ko', markersize=15) ...@@ -1163,21 +1148,16 @@ plt.plot(x,y,'ko', markersize=15)
plt.plot(x+1,y+1, 'o', markerfacecolor='none', markeredgecolor='k', markersize=15); plt.plot(x+1,y+1, 'o', markerfacecolor='none', markeredgecolor='k', markersize=15);
``` ```
#### Question 1. 1. Sketch the Wigner-Seitz unit cell and two other possible primitive unit cells of the crystal.
Sketch the Wigner-Seitz unit cell and two other possible primitive unit cells of the crystal. 2. If the distance between the filled circles is $a=0.28$ nm, what is the area of the primitive unit cell?
#### Question 2. What would be the area of the primitive unit cell if all empty and filled circles were identical?
If the distance between the filled circles is $a=0.28$ nm, what is the area of the primitive unit cell? 3. Write down a set of primitive lattice vectors and the corresponding basis for this crystal.
How would this area change if all the empty circles and the filled circles were identical? Would these primitive lattice vectors still be primitive if all empty and filled circles were identical? If not, identify a new primitive unit cell and the associated basis.
#### Question 3. 4. Imagine expanding the lattice into the perpendicular direction $z$ (out of the page). We can define a new three-dimensional crystal by considering a periodic structure in the $z$ direction, where the filled circles have been displaced by $\frac{a}{2}$ in both the $x$ and $y$ direction from the empty circles.
Write down one set of primitive lattice vectors and the basis for this crystal.
What happens to the number of elements in the basis if all empty and filled circles were identical?
#### Question 4.
Imagine expanding the lattice into the perpendicular direction $z$. We can define a new three-dimensional crystal by considering a periodic structure in the $z$ direction, where the filled circles have been displaced by $\frac{a}{2}$ in both the $x$ and $y$ direction from the empty circles.
The figure below shows the new arrangement of the atoms. The figure below shows the new arrangement of the atoms.
What lattice do we obtain? What lattice do we obtain?
Write down the basis of the three-dimensional crystal. Write down the basis of this three-dimensional crystal.
#### Question 5. 5. If we consider all atoms to be the same, what lattice do we obtain?
If we consider all atoms to be the same, what lattice do we obtain?
Compute the filling factor in the case where all atoms are the same. Compute the filling factor in the case where all atoms are the same.
```python ```python
...@@ -1244,9 +1224,9 @@ layout=go.Layout(showlegend = False, ...@@ -1244,9 +1224,9 @@ layout=go.Layout(showlegend = False,
go.Figure(data=data, layout=layout) go.Figure(data=data, layout=layout)
``` ```
### Exercise 2: Diamond lattice ### Exercise 2: The crystal structure of diamond
Consider a the [diamond crystal structure](https://en.wikipedia.org/wiki/Diamond_cubic) structure. The following illustration shows the arrangement of the carbon atoms in a conventional unit cell. Consider the [diamond crystal structure](https://en.wikipedia.org/wiki/Diamond_cubic). The illustration shows the arrangement of the carbon atoms in a conventional unit cell.
```python ```python
Xn = np.tile(np.arange(0,2,1),4) Xn = np.tile(np.arange(0,2,1),4)
...@@ -1328,34 +1308,24 @@ go.Figure(data=data, layout=layout) ...@@ -1328,34 +1308,24 @@ go.Figure(data=data, layout=layout)
The side of the cube is $a = 0.3567$ nm. The side of the cube is $a = 0.3567$ nm.
#### Question 1. 1. How is this crystal structure related to the fcc lattice?
How is this crystal structure related to the fcc lattice? Write down a set of primitive lattice vectors and compute the volume of the corresponding primitive unit cell.
Write down one set of primitive lattice vectors and compute the volume of the corresponding primitive unit cell. 2. How many atoms are in the primitive unit cell? And how many lattice points? Write down the basis.
#### Question 2. 3. Determine the number of atoms in the conventional unit cell and compute this unit cell's volume.
How many atoms are in the primitive unit cell? 4. What is the distance between nearest-neighbour atoms?
Write down the basis. 5. Compute the atom density (expressed in nr of atoms per cubic nanometer) and the filling factor.
#### Question 3.
Determine the number of atoms in the conventional unit cell and compute its volume. ### Exercise 3: Directions and spacings of Miller planes
#### Question 4.
What is the distance between nearest neighbouring atoms?
#### Question 5.
Compute the filling factor.
### Exercise 3: Directions and Spacings of Miller planes
*(adapted from ex 13.3 of "The Oxford Solid State Basics" by S.Simon)*
#### Question 1. *(adapted from ex 13.3 of "The Oxford Solid State Basics" by S.Simon)*
Explain what is meant by the terms Miller planes and Miller indices.
#### Question 2.
Consider a cubic crystal with one atom in the basis and a set of orthogonal primitive lattice vectors $a\hat{x}$, $a\hat{y}$ and $a\hat{z}$. Show that the direction $[hkl]$ in this crystal is normal to the planes with Miller indices $(hkl)$.
#### Question 3.
Show that this is not true in general. Consider for instance an orthorhombic crystal, for which the primitive lattice vectors are still orthogonal but have different lengths.
#### Question 4.
Any set of Miller indices corresponds to a family of planes separated by a distance $d$. Show that the spacing $d$ of the $(hkl)$ set of planes in a cubic crystal with lattice parameter $a$ is $d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$.
??? hint 1. Explain the terms Miller planes and Miller indices.
2. Consider a cubic crystal with one atom in the basis and a set of orthogonal primitive lattice vectors $a\hat{x}$, $a\hat{y}$ and $a\hat{z}$. Show that the direction $[hkl]$ in this crystal is normal to the planes with Miller indices $(hkl)$.
3. Show that this is not true in general. Consider for instance an orthorhombic crystal, for which the primitive lattice vectors are still orthogonal but have different lengths.
4. Any set of Miller indices corresponds to a family of planes separated by a distance $d$. Show that the spacing $d$ of the $(hkl)$ set of planes in a cubic crystal with lattice parameter $a$ is $d = \frac{a}{\sqrt{h^2 + k^2 + l^2}}$.
Recall that a family of lattice planes is an infinite set of equally separated parallel planes which taken all together contain all points of the lattice. ??? hint
Try computing the distance between the plane that contains the site $(0,0,0)$ of the conventional unit cell and a plane defined by the $(hkl)$ indices. Recall that a family of lattice planes is an infinite set of equally separated parallel planes which taken all together contain all points of the lattice.
Try computing the distance between the plane that contains the site $(0,0,0)$ of the conventional unit cell and a plane defined by the $(hkl)$ indices.
docs/9_crystal_structure_solutions.md
+ 73
− 114
View file @ 95cdb563
...@@ -10,53 +10,44 @@ from math import pi ...@@ -10,53 +10,44 @@ from math import pi
# Solutions for lecture 9 exercises # Solutions for lecture 9 exercises
## Warm-up exercises ## Warm-up exercises
#### Question 1. 1. A primitive unit cell is a unit cell which contains a single lattice point. Because it only contains a lattice point, it is the unit cell with the smallest volume out of all unit cells.
A primitive unit cell is a unit cell which contains a single lattice point. 2. Since the primitive unit cell should contain one lattice point, its volume is $a^3/4$.
Because it only contains a lattice point, it is the unit cell with the smallest volume out of all unit cells.
#### Question 2. 3. Let us first consider the FCC lattice. Its primitive lattice vectors vectors are
Let us first consider the FCC lattice. \begin{align}
Its primitive lattice vectors vectors are \mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}})\\
\mathbf{a}_2 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{z}})\\
\mathbf{a}_3 &= \frac{a}{2}(\mathbf{\hat{y}} + \mathbf{\hat{z}}).
\end{align}
\begin{align} With respect to the conventional unit cell, the basis in fractional coordinates is $\bigcirc(1/2,1/2,0)$, $\bigcirc(1/2,0,1/2)$, $\bigcirc(0,1/2,1/2)$ and $\bigcirc(0,0,0)$.
\mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}})\\ With respect to the primitive unit cell, the basis is $\bigcirc(0,0,0)$.
\mathbf{a}_2 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{z}})\\ Let us now consider the BCC lattice.
\mathbf{a}_3 &= \frac{a}{2}(\mathbf{\hat{y}} + \mathbf{\hat{z}}). The primitive lattice vectors are
\end{align}
With respect to the conventional unit cell, the basis in fractional coordinates is $\bigcirc(1/2,1/2,0)$, $\bigcirc(1/2,0,1/2)$, $\bigcirc(0,1/2,1/2)$ and $\bigcirc(0,0,0)$. \begin{align}
With respect to the primitive unit cell, the basis is $\bigcirc(0,0,0)$. \mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}} + \mathbf{\hat{z}})\\
Let us now consider the BCC lattice. \mathbf{a}_2 &= a\mathbf{\hat{x}}\\
The primitive lattice vectors are \mathbf{a}_3 &= a\mathbf{\hat{y}}.
\end{align}
\begin{align} The basis of the conventional unit cell is $\bigcirc(0,0,0)$ and $\bigcirc(1/2,1/2,1/2)$.
\mathbf{a}_1 &= \frac{a}{2}(\mathbf{\hat{x}} + \mathbf{\hat{y}} + \mathbf{\hat{z}})\\ For the primitive unit cell the basis is $\bigcirc(0,0,0)$.
\mathbf{a}_2 &= a\mathbf{\hat{x}}\\
\mathbf{a}_3 &= a\mathbf{\hat{y}}.
\end{align}
The basis of the conventional unit cell is $\bigcirc(0,0,0)$ and $\bigcirc(1/2,1/2,1/2)$. 4. You would need at least two. A diatomic crystal could require more than two basis vectors if the atoms as determined by the spatial location of the atoms. One needs to be able to construct all atoms in the crystal using a basis plus a lattice.
For the primitive unit cell the basis is $\bigcirc(0,0,0)$.
#### Question 3. 5. The filling factor is
You would need at least two. A diatomic crystal could require more than two basis vectors if the proportion between the two elements is not one to one. $$
F = \frac{\pi}{6}
$$
#### Question 4. 6. See lecture notes
$$ ## Exercise 1: Diatomic crystal
F = \frac{\pi}{6}
$$
#### Question 5.
See lecture notes
## Exercise 1: Diatomic crystal¶
#### Question 1.
```python ```python
y = np.repeat(np.arange(0,8,2),4) y = np.repeat(np.arange(0,8,2),4)
...@@ -77,110 +68,78 @@ plt.plot([6,4,2,4,6],[0,0,2,2,0], color='k',ls=':') ...@@ -77,110 +68,78 @@ plt.plot([6,4,2,4,6],[0,0,2,2,0], color='k',ls=':')
plt.plot(x,y,'ko', markersize=15) plt.plot(x,y,'ko', markersize=15)
plt.plot(x+1,y+1, 'o', markerfacecolor='none', markeredgecolor='k', markersize=15); plt.plot(x+1,y+1, 'o', markerfacecolor='none', markeredgecolor='k', markersize=15);
``` ```
#### Question 2.
The area of the primitive unit cell is $A = a^2$.
If the filled and empty circles are identical particle, the nearest neighbour distance becomes $a^* = \frac{a}{\sqrt{2}}$ and thus the area $A^* = {a^*}^2 = \frac{a^2}{2} = \frac{A}{2}$.
#### Question 3. 1. See plot above
One set of primitive lattice vectors is 2. The area of the primitive unit cell is $A = a^2$.
$$ If the filled and empty circles are identical particles, the nearest-neighbour distance becomes $a^* = \frac{a}{\sqrt{2}}$ and thus the area $A^* = {a^*}^2 = \frac{a^2}{2} = \frac{A}{2}$.
\mathbf{a_1} = a \hat{\mathbf{x}}, \quad \mathbf{a_2} = a \hat{\mathbf{y}}.
$$
With respect to the primitive lattice vectors, the basis is
$
\huge \bullet \normalsize(0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2}).
$
#### Question 4. 3. One set of primitive lattice vectors is
The lattice is a cubic lattice. $$
The basis of the crystal is \mathbf{a_1} = a \hat{\mathbf{x}}, \quad \mathbf{a_2} = a \hat{\mathbf{y}}.
$$
$ With respect to the primitive lattice vectors, the basis is
\huge \bullet \normalsize(0,0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2},\frac{1}{2}).
$
An example of such a material is Cesium Chloride (CsCl) $$
\huge \bullet \normalsize(0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2}).
$$
#### Question 5. 4. The lattice is a cubic lattice. The basis of the crystal is
We obtain the BCC lattice. $$
\huge \bullet \normalsize(0,0,0), \quad \bigcirc(\frac{1}{2},\frac{1}{2},\frac{1}{2}).
Example: Sodium (Na) $$
#### Question 6.
$$
F = \frac{\sqrt(3)\pi}{8}
$$
## Exercise 2: Diamond lattice
#### Question 1.
The conevtional unit cell of diamond consists out of two shifted fcc lattices. An example of such a material is Cesium Chloride (CsCl)
One set of primitive lattice vectors is
\begin{align}
\mathbf{a_1} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right) \\
\mathbf{a_2} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_3} &= \frac{a}{2} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right).
\end{align}
The volume of the primitive unit cell is
$$
V = \left| \mathbf{a_1} \cdot \left(\mathbf{a_2} \times \mathbf{a_3} \right) \right| = \frac{a^3}{4}
$$
#### Question 2. 5. We obtain the BCC lattice. Example: Sodium (Na)
The primitive unit cell contains 2 atoms. 6. The filling factor is
With respect to the set of primitive lattcie vectors, the basis is $ \mathrm{C}(0,0,0)$ and $\mathrm{C}(\frac{1}{4},\frac{1}{4},\frac{1}{4})$.
#### Question 3. $$
F = \frac{\sqrt(3)\pi}{8}
$$
One FCC lattice contains 4 atom. ## Exercise 2: Diamond lattice
Because the diamond conventional unit cell contains two shifted FCC lattices, it will contain 8 atoms.
The volume of the conventional unit cell is $V = a^3$.
#### Question 4.
??? hint "Visual hint"
Consider the atom at (0.25,0.25,0.25) coordinates in the interactive diamond lattice image and look at its neighbours.
The distance between neirest neighbouring atoms is $d = \frac{\sqrt{3}a}{4}$ 1. The conventional unit cell of diamond consists of two intertwined fcc lattices. A possible set of primitive lattice vectors is
#### Question 5. \begin{align}
\mathbf{a_1} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{y}} \right) \\
\mathbf{a_2} &= \frac{a}{2} \left(\hat{\mathbf{x}}+\hat{\mathbf{z}} \right) \\
\mathbf{a_3} &= \frac{a}{2} \left(\hat{\mathbf{y}}+\hat{\mathbf{z}} \right).
\end{align}
$$ The volume of the primitive unit cell is
F = \frac{\sqrt{3}\pi}{16}
$$
## Exercise 3: Directions and Spacings of Miller planes $$
#### Question 1. V = \left| \mathbf{a_1} \cdot \left(\mathbf{a_2} \times \mathbf{a_3} \right) \right| = \frac{a^3}{4}
$$
Miller plane is a plane that intersects an infinite number of lattice points. as expected because the conventional unit cell contains 4 lattice points.
Miller indeces are a set of 3 integers which specify a set of parallel planes.
#### Question 2. 2. The primitive unit cell contains 2 atoms. With respect to the set of primitive lattice vectors, the basis is $ \mathrm{C}(0,0,0)$ and $\mathrm{C}(\frac{1}{4},\frac{1}{4},\frac{1}{4})$.
??? hint "Small hint" 3. The conventional unit cell of the FCC lattice contains 4 atoms.
Because the diamond conventional unit cell contains two shifted FCC lattices, it contains 8 atoms.
The volume of the conventional unit cell is $V = a^3$.
The $(hkl)$ plane intersects lattice at position vectors of $\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_1}}{l}$. 4. We indentify a nearest neighbour pair consisting of the atom at the origin and the atom at (0.25a,0.25a,0.25a). Therefore the distance between neirest neighbouring atoms is $d = \frac{\sqrt{3}a}{4}$
Can you define two general vector inside the $(hkl)$ plane that span the plane?
#### Question 3. 5. The atom density is $\rho=N_\text{atom}/a^3 = 175/$ nm$^3$. The filling factor is
The same hint as mentioned in the previous subquestion also applies here. $$
F = \frac{\sqrt{3}\pi}{16}
$$
#### Question 4 ## Exercise 3: Directions and spacings of Miller planes
??? hint "Big hint" 1. A Miller plane is a plane that intersects an infinite number of lattice points. Miller indices are a set of 3 integers which specify a set of parallel planes.
There is always a neighbouring lattice plane which intersects the (0,0,0) lattice point. 2. The $(hkl)$ plane intersects the lattice vectors at position vectors of $\frac{\mathbf{a_1}}{h}, \frac{\mathbf{a_2}}{k}, \frac{\mathbf{a_1}}{l}$. Can you define two vectors inside the $(hkl)$ plane that span the plane? How do you compute a vector that is orthogonal to two other vectors?
??? hint "Small hint" 3. The procedure described in the previous subquestion also applies here.
Do not forget to normalize your unit vector. 4. We can calculate the distance between two planes by calculating the projection of any vector connecting the two planes onto a unit vector that is normal to the plane.
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